Work And Rates:

by on September 25th, 2009

There is one very important equation that guides all rate and work questions: r = d/t, “rate equals distance over time.” If given any two of the three, you should be able to find the third, if the units remain constant. Problems involving “work” are essentially rate problems, where a worker’s “efficiency” is calculated by amount completed in a given period of time.

Working Together

In questions where individuals work at different speeds, we typically need to add their separate rates together. Make sure you keep your units straight. This doesn’t mean wasting time and writing each and every one out, but rather simply recognizing their existence. Note that when working together, the total time to complete the same task will be less than BOTH of the individual rates, but not necessarily in proportion. Nor, are you averaging or adding the given times taken. You must add rates.

A worker can load 1 full truck in 6 hours. A second worker can load the same truck in 7 hours. If both workers load one truck simultaneously while maintaining their constant rates, approximately how long, in hours, will it take them to fill 1 truck?

A. 0.15

B. 0.31

C. 2.47

D. 3.23

E. 3.25

The rate of worker #1 is 1 truck/6 hours. This can also be 1/6 trucks/1 hour. The rate of worker #2 is 1/7. When together, they will complete 1/6 + 1/7 trucks/ 1 hour.

1/6 + 1/7 = 6/42 + 7/42 = 13/42 trucks/1 hour. Remember the question is asking for the number of hours to fill 1 truck, NOT the number of trucks completed in 1 hour. To find this, we find the reciprocal of 13/42.

42/13 hours/truck = 3 3/13 hours/truck.

At this point, we may not be able to decide between (D) or (E). However, the decimal is important. Because the denominator is 13, we know the decimal cannot equal .25. We can also see that 3/12 will yield .25, so 3/13 will be slightly lower. Choice (E).

Relative Velocity

Planes, trains and automobiles. Sometimes walking. Objects moving at given speeds on the GMAT usually travel toward or away from each other. When moving at an angle, we may be looking at a geometry question. If moving toward or away from each other, we can add their speeds to see their relative velocities. If moving in the same direction, we instead subtract their speeds to find the relative velocity. Again, be careful of units.

Train A traveling at 60 m/hr leaves New York for Dallas at 6 P.M. Train B traveling at 90 m/hr also leaves New York for Dallas at 9 P.M. Train C leaves Dallas for New York at 9 P.M. If all three trains meet at the same time between New York and Dallas, what is the speed of Train C if the distance between Dallas and New York is 1260 miles?

A. 60 m/hr

B. 90 m/hr

C. 120 m/hr

D. 135 m/hr

E. 180 m/hr

Relative to Train A, Train B’s velocity is 30 m/hr. This means that B will gain on A at a rate of 30 miles every hour. In the three hours from 6pm to 9pm, A gets to mile marker 180. To catch up the 180 miles, it will take Train B 6 hours. So when they all meet up, the time will be 3am, and they will be at mile marker 540. Notice that:

Train A = 9 hours at 60 miles/hour = 540 miles

Train B = 6 hours at 90 miles/hour = 540 miles

We can now tackle Train C, which has traveled the same time as B (6 hours), and traveled (1260 – 540) miles.

Rate of Train C = 720 miles/ 6 hours = 120 miles/hour. Choice (C).

Man Hours

Many times you may be asked to calculate the number of workers would be need to complete a certain task. Keep in mind that the number of workers (at the same efficiency) is inversely proportional to the amount of time it takes one to complete a given task. It may help consider the unit man-hours as the multiplication between workers and time, which is then compared to the work completed. For example:

Three plows working at identical constant rates can clear 123 ft of snow per minute. At this rate, how much snow could 8 plows remove in 5 minutes?

A. 328

B. 984

C. 1,640

D. 16,400

E. 131,200

Instead of man-hours, here we want to interact plow-minutes. Feet and minutes are already compared, so all we have to is add “plows” to the expression. If we divide 123 ft/min by 3 plows, we get:

123 ft/minute/3 plows = 41 ft/plow-minute

At this rate, if we want to increase minutes to 5 and plows to 8, we can simply insert these into the existing rate. Note the absolute rate does not change, since we are multiplying top and bottom by 40, so the value is constant.

41*40 feet / 40 plow-minutes = 1640 feet / 40 plow-minutes. Choice (C).

There are LOADS more rate questions; some are much more difficult. This is by no means exhaustive. Post some tough ones and we can tackle them together!


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